2 条题解

  • 2
    @ 2022-11-13 13:32:14
    #include <bits/stdc++.h>
    
    using namespace std;
    
    int main()
    {
        int a;
    
        cin >> a;
    
        if (a % 2 == 0)
        {
            cout << "even";
        }
        else
        {
            cout << "odd";
        }
    
        return 0;
    }
    
    • 1
      @ 2025-3-11 13:46:24
      #include <queue>
      #include <math.h>
      #include <stack>
      #include <stdio.h>
      #include <iostream>
      #include <vector>
      #include <iomanip>
      #include <string.h>
      #include <algorithm>
      using namespace std;
      #define elif else if
      #define int long long
      #define float double
      #define N 30000010
      #define INF 0xc00000fd
      #define scf scanf
      #define ptf printf
      #define gtc getchar
      #define ptc putchar
      #define inr int n = read()
      #define imr int m = read()
      #define psp ptc(' ')
      #define pel ptc('\n')
      #define pzr ptc('0')
      #define fin for(int i = 1; i <= n; i++)
      #define fim for(int i = 1; i <= m; i++)
      #define fjn for(int j = 1; j <= n; j++)
      #define fjm for(int j = 1; j <= m; j++)
      #define rai fin	a[i] = read()
      #define wct write(cnt)
      #define pct print(cnt)
      #define was write(ans)
      #define pas print(ans)
      queue<int> Q;
      stack<int> S;
      vector<int> V;
      inline int wei(int n){
      	int cnt = 0;
      	while(n > 0){
      		n /= 10;
      		cnt++;
      	}
      	return cnt;
      }
      inline bool isPrime(int n){
      	if(n < 2)
      		return 0;
      	for(int i = 2; i * i <= n; ++i)
      		if(n % i == 0)
      			return 0;
      	return 1;
      }
      inline int read(string n){
      	int x = 0, f = 1;
      	ptf("%s", n.c_str());
      	char c = gtc();
      	while(c < '0'  ||  c > '9'){
      		if(c == '-')
      			f = -1;
      		c = gtc();
      	}
      	while(c >= '0'  &&  c <= '9'){
      		x = x * 10 + c - 48;
      		c = gtc();
      	}
      	return x * f;
      }
      inline float input(string n){
      	float x = 0, f = 1, x2 = 0, cnt = 0, i = 0;
      	ptf("%s", n.c_str());
      	char c = gtc();
      	while(c < '0'  ||  c > '9'){
      		if(c == '-')
      			f = -1;
      		c = gtc();
      	}
      	while(c >= '0'  &&  c <= '9'){
      		x = x * 10 + c - 48;
      		c = gtc();
      	}
      	c = gtc();
      	while(c >= '0'  &&  c <= '9'){
      		x2 = x2 * 10 + c - 48;
      		cnt++;
      		c = gtc();
      	}
      	for(; i < cnt; i++)
      		x2 /= 10.0;
      	return (x + x2) * f;
      }
      inline void write(int n) {
          if(n < 0){
              ptc('-');
              n = -n;
          }
          if(n > 9)
      		write(n / 10);
          ptc(n % 10 + '0');
      	return;
      }
      inline void print(int n, float m){
      	write(m);
      	m -= (int)m;
      	ptc('.');
      	int cnt = 0;
      	while(m - (int)m && n - cnt){
      		m *= 10;
      		ptc((int)m % 10 + '0');
      		cnt++;
      	}
      	n -= cnt;
      	fin
      		pzr;
      	return;
      }
      #define input() input("")
      #define read() read("")
      inr;
      inline void Main(){
      	puts(n % 2 ? "odd" : "even");
      	return;
      }
      signed main(signed argc, char **argv){
      	Main();
      	pel;
      	return 0;
      }
      
      • 1

      信息

      ID
      869
      时间
      1000ms
      内存
      128MiB
      难度
      3
      标签
      递交数
      109
      已通过
      61
      上传者